An optical fiber consists of core of `mu_(1)` surrounded by a cladding of `mu_(2) lt mu_(1)`. A beam of light enters from air at an angle `alpha` with axis of fiber. The highest `alpha` for which ray can be travelled through fiber is

Options

A. `cos^(-1)sqrt(mu_(2)^(2)-mu_(1)^(2))`
B. `sin^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`
C. `tan^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`
D. `sec^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`

Answer

B

Explanation

Here the requirement is that `igtc`
`impliessinigtsincimpliessinigt(mu_(2))/(mu_(1))` ...`(i)`
From Snell's law `mu_(1)=(sin alpha)/(sin r)` ...`(ii)`
Also in `DeltaOBA`
`r+i=90^(@)impliesr=(90-i)`
Hence from equation `(ii)`
`sin alpha=mu_(1)sin (90-i)`
`impliescosi=(sin alpha)/(mu_(1))`
`sin i =sqrt(1-cos^(2)i)=sqrt(1((sinalpha)/(mu_(1)))^(2))`
From equations (i) and (iii) `sqrt(1-((sin alpha)/(mu_(1)))^(2)) gt (mu_(2))/(mu_(1))`
`rArr sin^(2)alpha lt (mu_(1)^(2)-mu_(2)^(2)) rArr sin alpha lt sqrt(mu_(1)^(2)-mu_(2)^(2))`
`alpha_(max)=sin^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`